Lab 13B: Solubility

Introduction: The purpose of this lab was to find an unknown substance (given three options-NaCl, KCl, and NaNO3) by creating a method - involving solubility, which is a term for how soluble a solute (the substance you're trying to dissolve) is in a solvent (the substance in which you try and dissolve a solute) - to do so. How we'd know which solute was the one we were given was provided to us (indirectly) on a graph, which charted the solubility for each of the three solutes given. So, as a result, the core of this lab was to find a substance by making a procedure, when only given a few materials and our own minds.

Materials: 2 beakers of different sizes, a plastic tray, the substance, a graduated cylinder, about 10mL of water, a stirring rod (to be used to stir and break up the solid), a thermometer, and a hot plate.

Procedure: First, we filled up the smaller of the two beakers with water, and found its temperature. Then, we obtained a plastic tray and found the mass of it, then poured the solid in and found the mass of the solid and the tray combined, then subtracted the combined mass by the mass of the tray alone to find the mass of the substance. We pour the substance into the room-temperature beaker (the other beaker is currently sitting on the hot plate heating up some water so we can heat the other beaker indirectly by putting the small beaker into the big beaker), and try and dissolve it for 2 minutes. As the substance is put into a temperature where, on the graph, two solutes have similar solubility points, and one's solubility point given that temperature is very far away from the others. If the solute dissolves, it's the one that's far away, but if it doesn't, it's one of the two that are very close to each other given the temperature. Unfortunately, it doesn't dissolve completely, and we have to use the heated beaker. We place the small beaker inside the big beaker, and heat up the water to the point where the two previously-close substances (in terms of solubility) now have far-apart solubility rates given a certain temperature. If the substance dissolves, it's Potassium Nitrate, but if it doesn't, it's Sodium Chloride (which makes the previously-eliminated substance Sodium Nitrate). Luckily, it dissolves, so the substance is found to be Potassium Nitrate.

Data:

Temperatures of water:
- 20.5 degrees Celsius (small beaker at room temperature).
- 45.6 degrees Celsius (small beaker inside the larger beaker).

Masses:
- 2.04g (tray)
- 7.44g (tray + solid)
- 5.40g (solid).

Volume of Water: 10mL.

Conclusion: The solid is Potassium Nitrate, as a) it didn't dissolve at 20.5 degrees Celsius (which it would've if it was Sodium Nitrate), and b) it did dissolve at 45.6 degrees Celsius (which it wouldn't have if it was Sodium Chloride). This also shows the relationship between temperature and solubility, which is that if one increases, the other will increase as well.